Find $y'$.
$$ \\ \begin{align} \\5{y}^2 &= \dfrac{2x - 3}{2x + 3} \\ \\5{y}^2(2x + 3) &= 2x - 3 \\ \\ y(2x+3)&= \dfrac{2x-3}{5y} \\ \\ \frac{\partial y}{\partial x}[y(2x + 3)] &= \frac{\partial y}{\partial x}[\dfrac{2x - 3}{5y}] \\ \\ y'(2x + 3) + 2y &= \dfrac{10y - 5y'(2x - 3)}{{(5y)}^2} \\ \\ 2xy'+ 3y'+ 2y &= \dfrac{10y - 10xy' + 15y'}{25{y}^2} \\ \\ 50x{y}^2y' + 75{y}^2y' + 50{y}^3 &= 10y - 10xy' + 15y' \\ \\ 5(10x{y}^2y' + 15{y}^2y' + 10y^3) &= 5(2y - 2xy' + 3y') \\ \\ 10x{y}^2y' + 15{y}^2y' + 10{y}^3 &= 2y - 2xy' + 3y' \\ \\ 10x{y}^2y' + 15{y}^2y' + 2xy' - 3y' &= 2y - 10{y}^3 \\ \\ y'(10x{y}^2 + 15{y}^2 + 2x - 3) &= 2y(1-5{y}^2) \\ \\ y' &= \dfrac{2y(1-5{y}^2)}{10x{y}^2 + 15{y}^2 + 2x - 3} \\ \end{align} $$
I find $y'$ to be $y' = \dfrac{2y(1-5{y}^2)}{10x{y}^2 + 15{y}^2 + 2x - 3}$.
Could someone verify or dispute this?